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2 years ago

How is capacitance related to the distance between the plates of a capacitor?

Physics

1 answer:

svetlana [45]2 years ago

5 0

Answer:C

Explanation: I studied, and C is correct

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An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

1.228 x $10^{-6}$ mm

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = 1.228 x $10^{-6}$ mm

6 0

2 years ago

Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E i.e

$\Delta K = \Delta V$

If the initial and final velocity of the proton is $u$ and $v$ respectively then,

change in Kinetic Energy $\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0$

change in Potential Energy $\implies \Delta U = q\Delta V$

from conservation of energy,

$v= \sqrt{\frac{2q\Delta V}{m}}$

so, $v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}$

$= 25.86 \times 10^4 ~m/s$

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0

1 year ago

Which rf characteristic best determines the range of a 2.4 ghz ism signal?

Rainbow [258]

The radio frequency characteristic that best determines the range of a 2.4 GHz ism signal is the wavelength.

This frequency can be used in WiFi and can reach up to 46 meters when indoors and about 92 meters when outdoors.

3 0

3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

LC Circuit

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

2 years ago

Please help, it's grade 7 science. :)

xeze [42]

Answer:

The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.

The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.

4 0

2 years ago

Read 2 more answers